Question

(х+1)(х+2)(х+3)(х+4)=(х+1)(х+2)(х+3)(х+5) 

Answer 1

(х+1)(х+2)(х+3)(х+4)-(х+1)(х+2)(х+3)(х+5)=0
(х+1)(х+2)(х+3)(х+4-х-5)=0
(х+1)(х+2)(х+3)*(-1)=0
произведение равно нулю когда один из множителей равен нулю т.е.
х+1=0 или х+2=0 или х+3=0
х₁=-1, х₂=-2, х₃=-3

Answer 2

Expanding both sides of the equation, we get:

(х+1)(х+2)(х+3)(х+4) = (х+1)(х+2)(х+3)(х+5)

Multiplying out the terms on both sides, we get:

x^4 + 10x^3 + 35x^2 + 50x + 24 = x^4 + 10x^3 + 35x^2 + 50x + 30

Subtracting x^4, 10x^3, 35x^2, and 50x from both sides, we get:

24 = 30

This is a contradiction, which means that there are no solutions to the equation. Therefore, the statement (х+1)(х+2)(х+3)(х+4)=(х+1)(х+2)(х+3)(х+5) is false for all values of x.